The sampling operator \(T\)  is one-to-one on \(S_\mathcal {D}(\mathcal {A}^{\prime })\)  if and only if for all \(t \in \mathcal {M} \text{ and } i \in I, <f, M_{g_t}\psi _i> = 0\)  will imply \(f=0\)  for any \(f \in S_\mathcal {D}(\mathcal {A}^{\prime })\)  which is equivalent to \((ii).\) 
Therefore statements (i) and (ii) are equivalent. Similarly, statements (iii) and (iv) are equivalent. Also, if (ii) is true, then by [1]}, the corresponding range function \(J_{\mathcal {A}^{\prime }}\)  satisfies
\(J_{\mathcal {A}^{\prime }} (x)=\overline{\operatornamewithlimits{span}} \lbrace \psi _i(x) : i \in I\rbrace .\) 
Since \(J_{\mathcal {A}^{\prime }}(x)\)  is closed, then by [2]}, \(\tilde{T}(x)\)  is one-to-one on \(J_{\mathcal {A}^{\prime }}(x)\)  for a.e. \(x \in X\)  and hence the statements (ii) and (iii)
are equivalent. The equivalence of the statements (i), (v), and (vi) follows from the fact that the fiber space \(J_{\mathcal {A}^{\prime }}(x)\)  is finite dimensional and the range of the synthesis operator \(T_{\mathcal {A}^{\prime }}^*(x)\)  is \(J_{\mathcal {A}^{\prime }}(x)\) .
